How Does Integral Rewriting Transform Two Integrals into One?

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On the attached picture the double integral in the first line is rewritten in the second line by introducting the variable τ=τ1-τ2
But how exactly does this happen? I simply can't see how two integrals can turn into one.
 

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The first integral is just beta. It eventually cancels out.
 
I don't understand why it is just beta. There is a Green's function in the integrand. Do you say this because there is no beta in the second line? Because for all I know this could be a typo.
 
Can you do the full substitution where you introduce new variables, say
\tau = \tau_1 - \tau_2 \text{ and } \mu = \tau_1 + \tau_2
to turn a two dimensional integral into a two dimensional integral? You will see the integration over \mu become trivial.
 
hmm I'd would say it would give something like the attached, but I don't see any trivial integral. What did I do wrong?
 

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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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